Problem 18 - Precalculus: Functions and Graphs

Question

Solve the system. $$\left\\{\begin{aligned} 3 x+7 y &=9 \\ y &=5 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{-26}{3}\), \(y = 5\).

Step 1: Identify the System of Equations

We have a system of two linear equations: 1. \(3x + 7y = 9\)2. \(y = 5\)

Step 2: Substitute 'y' Value

Substitute \(y = 5\) from the second equation into the first equation. The first equation \(3x + 7y = 9\) becomes:\(3x + 7(5) = 9\).

Step 3: Simplify the Equation

Calculate the value of \(7 \times 5\) to replace in the equation: \(3x + 35 = 9\).

Step 4: Solve for 'x'

Isolate \(x\) by subtracting 35 from both sides:\(3x = 9 - 35\)\(3x = -26\).Next, divide by 3 to solve for \(x\):\(x = \frac{-26}{3}\).

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines when plotted on a graph. These equations involve variables raised only to the power of one. They take the general form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. In our system:

• First equation: \(3x + 7y = 9\)
• Second equation: \(y = 5\)

Linear equations are fundamental because they model many real-world situations such as finances, physics, and statistics. Understanding how to solve linear equations helps in developing important problem-solving skills. Solving these equations often involves manipulating them to find unknown values that satisfy all equations in a system.

Substitution Method

The substitution method is a powerful algebraic technique used to solve systems of equations. When using this method, you solve one equation for one variable and then substitute this expression into another equation. This helps reduce the system to a single equation with one variable.
In our example, the second linear equation is \(y = 5\). Here, \(y\) is already expressed as a constant. So, you can directly substitute this value into the first equation. Replace \(y\) in the equation \(3x + 7y = 9\) with 5:

• This turns the equation into \(3x + 7(5) = 9\).

This strategic substitution allows you to solve for the remaining variable, \(x\), in the next steps.

Solving for x

After substitution, the next step is solving for \(x\). Once the equation is simplified, it becomes a straightforward problem of finding the value of \(x\). With our equation \(3x + 35 = 9\), the steps are:

• Subtract 35 from both sides of the equation to isolate terms involving \(x\): \(3x = 9 - 35\).
• Simplify the arithmetic operation: \(3x = -26\).
• Divide every term by the coefficient of \(x\), which is 3, to solve for \(x\): \(x = \frac{-26}{3}\).

Solving for \(x\) in this manner is crucial as it provides the solution that satisfies the system of linear equations. Each step uses basic arithmetic to gradually isolate \(x\). After these calculations, you have determined \(x\) efficiently.

Isolation of Variables

Isolation of variables is an essential skill when solving equations. This means rearranging the equation to get a specific variable on one side of the equation. The goal is to simplify and manipulate the equation to reveal the value of the unknown variable clearly.
In our system, once \(y = 5\) was substituted, isolating \(x\) became essential. You isolated \(x\) in the equation \(3x + 35 = 9\) through:

• Moving the constant we found, 35, from the left side to the right: \(3x = 9 - 35\).
• Ensuring \(x\) is by itself by dividing by its coefficient: \(x = \frac{-26}{3}\).

Isolation helps in simplifying complex systems into comprehensible steps, making it easier to get the solution without errors. Therefore, mastering this skill will aid in solving even more complicated algebraic problems efficiently.